SOLUTION: Building on question 95079, I have the part that goes like this:The only part of the height equation that we have not discussed is the constant. You have probably noticed that the

Algebra ->  Length-and-distance -> SOLUTION: Building on question 95079, I have the part that goes like this:The only part of the height equation that we have not discussed is the constant. You have probably noticed that the      Log On


   

Question 102209: Building on question 95079, I have the part that goes like this:The only part of the height equation that we have not discussed is the constant. You have
probably noticed that the constant is always equal to the initial height of the ball (80 m in
our previous exercises). Now, let’s have you develop a height equation.
A ball is thrown upward from the roof of a building 100 m tall with an initial velocity
of 20 m/s. Use this information for exercises 35 to 38.
35. Science and medicine. Write the equation for the height of the ball.
36. Science and medicine. When will the ball reach a height of 80 m?
I am trying to solve #36. So far I have:
h=-5t^2+20t+100
0=t^2-4-20
0= 4+or-sqrt-5^2-4*1*-20/2*1
=4+or-sqrt16+80/2
4=+or-sqrt96/2
4+9.8/2=13.8/2=6.9
4-9.8/2=5.8/2=2.9
am I anywhere near where I need to be?
Thank you so much.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
h=-5t%5E2%2B20t%2B100 Start with the given equation (assuming you did everything right for problem 35)

When the height is 80, this means h=80

80=-5t%5E2%2B20t%2B100 So plug in h=80



0=-5t%5E2%2B20t%2B20 Subtract 80 from both sides


Let's use the quadratic formula to solve for t:


Starting with the general quadratic

at%5E2%2Bbt%2Bc=0

the general solution using the quadratic equation is:

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve -5%2At%5E2%2B20%2At%2B20=0 ( notice a=-5, b=20, and c=20)




t+=+%28-20+%2B-+sqrt%28+%2820%29%5E2-4%2A-5%2A20+%29%29%2F%282%2A-5%29 Plug in a=-5, b=20, and c=20



t+=+%28-20+%2B-+sqrt%28+400-4%2A-5%2A20+%29%29%2F%282%2A-5%29 Square 20 to get 400



t+=+%28-20+%2B-+sqrt%28+400%2B400+%29%29%2F%282%2A-5%29 Multiply -4%2A20%2A-5 to get 400



t+=+%28-20+%2B-+sqrt%28+800+%29%29%2F%282%2A-5%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-20+%2B-+20%2Asqrt%282%29%29%2F%282%2A-5%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



t+=+%28-20+%2B-+20%2Asqrt%282%29%29%2F-10 Multiply 2 and -5 to get -10

So now the expression breaks down into two parts

t+=+%28-20+%2B+20%2Asqrt%282%29%29%2F-10 or t+=+%28-20+-+20%2Asqrt%282%29%29%2F-10


Now break up the fraction


t=-20%2F-10%2B20%2Asqrt%282%29%2F-10 or t=-20%2F-10-20%2Asqrt%282%29%2F-10


Simplify


t=2-2%2Asqrt%282%29 or t=2%2B2%2Asqrt%282%29


So these expressions approximate to

t=-0.82842712474619 or t=4.82842712474619


So our possible solutions are:
t=-0.82842712474619 or t=4.82842712474619



However, since a negative time doesn't make sense, our only solution is t=4.82842712474619


So at about 4.8 seconds the ball will be 80 m high